Question

Tangents drawn from $P (1,8)$ to the circle $x^{2}+y^{2}-6 x-4 y$ $-11=0$ touches the circle at the points A and B, respectively. The radius of the circle which passes through the points of intersection of circles $x^{2}+y^{2}-2 x+6 y-6=0$ and $x^{2}+y^{2}-2 x-6 y+6=0$ and intersects the circumcircle of the $\Delta P A B$ orthogonally is equal to

• A. $\frac{\sqrt{73}}{4}$
• B. $\frac{\sqrt{71}}{2}$
• C. 3
• D. 2

Answer 1

Answer: (A)

Equation of circle circumscribing $\Delta P A B$ is:
$(x-1)(x-3)+(y-8)(y-2)=0 $
$\Rightarrow x^{2}+y^{2}-4 x-10 y+19=0$
Equation of circle passing through points of intersection of circles $x^{2}+y^{2}-2 x-6 y+6=0$ and $x^{2}+$ $y^{2}-2 x-6 y+6=0$ is given by
$\left(x^{2}+y^{2}-2 x-6 y+6\right)+\lambda\left(x^{2}+y^{2}-2 x-6 y-6\right)=0 $
$\Rightarrow x^{2}+y^{2}+\frac{(2 \lambda-2)}{\lambda+1} x-6 y+6$
As circle (ii) is orthogonal to circle (i), we have
$-2\left(\frac{2 \lambda-2}{\lambda+1}\right)-5(-6)=19+6$
$\Rightarrow 4 \lambda-4=5 \lambda+5 $
$\Rightarrow \lambda=-9$
Hence, required equation fo circle is:
$x^{2}+y^{2}+\frac{5}{2} x-6 y+6=0$
$\therefore $ Radius of circle $=\sqrt{\frac{25}{16}+9-6}=\frac{\sqrt{73}}{4}$